Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
vector<T>& operator--(vector<T>& v) {
  for (auto& i : v) --i;
  return v;
}
template <typename T>
vector<T>& operator++(vector<T>& v) {
  for (auto& i : v) ++i;
  return v;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
  for (auto& i : v) os << i << ' ';
  return os;
}
template <typename T, typename U>
istream& operator>>(istream& is, pair<T, U>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <typename T, typename U>
pair<T, U> operator-(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <typename T, typename U>
pair<T, U> operator+(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <typename T, typename U>
void umin(T& a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
void umax(T& a, U b) {
  if (a < b) a = b;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<vector<int>> v(n, vector<int>(m));
  cin >> v;
  int maxv = (*min_element((v[0]).begin(), (v[0]).end()));
  int ans = 1e9;
  vector<pair<string, pair<int, int>>> best;
  for (int v0 = 0; v0 <= maxv; ++v0) {
    vector<pair<string, pair<int, int>>> curpath;
    int cur = 0;
    auto u = v;
    bool ok = true;
    curpath.emplace_back("row", make_pair(1, v0));
    cur += abs(v0);
    for (int i = 0; i < m; ++i) {
      u[0][i] -= v0;
    }
    for (int i = 0; i < m; ++i) {
      int k = u[0][i];
      if (k < 0) ok = false;
      curpath.emplace_back("col", make_pair(i + 1, k));
      cur += abs(k);
      for (int j = 0; j < n; ++j) {
        u[j][i] -= k;
      }
    }
    if (v0 == 2) {
      42;
      ;
      42;
      ;
    }
    for (int i = 1; i < n; ++i) {
      int k = u[i][0];
      if (k < 0) ok = false;
      curpath.emplace_back("row", make_pair(i + 1, k));
      cur += abs(k);
      for (int j = 0; j < m; ++j) u[i][j] -= k;
    }
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j)
        if (u[i][j] != 0) ok = false;
    if (ok && ans > cur) {
      ans = min(ans, cur);
      best = curpath;
    }
  }
  if (ans == 1e9) ans = -1;
  cout << ans << '\n';
  for (auto [s, uv] : best) {
    for (int i = 0; i < uv.second; ++i) {
      cout << s << ' ' << uv.first << '\n';
    }
  }
  return 0;
}